熊猫连接产生南价值

我很好奇，为什么熊猫身上只有两个简单的数据框架:

shape: (66441, 1)
dtypes: prediction    int64
dtype: object
isnull().sum(): prediction    0
dtype: int64


shape: (66441, 1)
CUSTOMER_ID    int64
dtype: object
isnull().sum() CUSTOMER_ID    0
dtype: int64

形状相同，两者都没有 NaN 值

foo = pd.concat([initId, ypred], join='outer', axis=1)
print(foo.shape)
print(foo.isnull().sum())

如果连接，就会产生大量的 NaN 值。

(83384, 2)
CUSTOMER_ID    16943
prediction     16943

我如何解决这个问题并防止 NaN 值被引入？

试图把它复制成

aaa  = pd.DataFrame([0,1,0,1,0,0], columns=['prediction'])
print(aaa)
bbb  = pd.DataFrame([0,0,1,0,1,1], columns=['groundTruth'])
print(bbb)
pd.concat([aaa, bbb], axis=1)

例如，由于没有引入 NaN 值，因此可以正常工作。

99483

小开

最佳答案

I think there is problem with different index values, so where concat cannot align get NaN:

aaa  = pd.DataFrame([0,1,0,1,0,0], columns=['prediction'], index=[4,5,8,7,10,12])
print(aaa)
prediction
4            0
5            1
8            0
7            1
10           0
12           0


bbb  = pd.DataFrame([0,0,1,0,1,1], columns=['groundTruth'])
print(bbb)
groundTruth
0            0
1            0
2            1
3            0
4            1
5            1


print (pd.concat([aaa, bbb], axis=1))
prediction  groundTruth
0          NaN          0.0
1          NaN          0.0
2          NaN          1.0
3          NaN          0.0
4          0.0          1.0
5          1.0          1.0
7          1.0          NaN
8          0.0          NaN
10         0.0          NaN
12         0.0          NaN

Solution is reset_index if indexes values are not necessary:

aaa.reset_index(drop=True, inplace=True)
bbb.reset_index(drop=True, inplace=True)


print(aaa)
prediction
0           0
1           1
2           0
3           1
4           0
5           0


print(bbb)
groundTruth
0            0
1            0
2            1
3            0
4            1
5            1


print (pd.concat([aaa, bbb], axis=1))
prediction  groundTruth
0           0            0
1           1            0
2           0            1
3           1            0
4           0            1
5           0            1

EDIT: If need same index like aaa and length of DataFrames is same use:

bbb.index = aaa.index
print (pd.concat([aaa, bbb], axis=1))
prediction  groundTruth
4            0            0
5            1            0
8            0            1
7            1            0
10           0            1
12           0            1

concatenated_dataframes = concat( [ dataframe_1.reset_index(drop=True), dataframe_2.reset_index(drop=True), dataframe_3.reset_index(drop=True) ], axis=1, ignore_index=True, ) concatenated_dataframes_columns = [ list(dataframe_1.columns), list(dataframe_2.columns), list(dataframe_3.columns) ] flatten = lambda nested_lists: [item for sublist in nested_lists for item in sublist] concatenated_dataframes.columns = flatten(concatenated_dataframes_columns)